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3 years ago

Is y=-x^2+4x a linear function?

Mathematics

2 answers:

andrey2020 [161]3 years ago

7 0

Answer:

It is not a linear function but it is a quadratic function of the type: y=ax²+bx+c.

Roman55 [17]3 years ago

5 0

No. That square of the variable makes it nonlinear. Its graph is a parabola.

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Simplify. 8r(4 - p) 32r -p 32r - 8rp 32 - rp 4 - 8rp

gizmo_the_mogwai [7]

So the right answer is of option B.

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Find the measure of ∠SMK. Please show your work, so I know the process.

Dahasolnce [82]

Answer: SMK is 66

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3 years ago

the original price of a DVD is $11. The sale price is 30% off the original price. what is the sale price of the DVD?

Dmitry [639]

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What is the midpoint of (10,6) and (4,9)

Leona [35]

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3 years ago

A certain forest covers an area of

iVinArrow [24]

Answer:

2598 square kilometers

Step-by-step explanation:

Hello

Step 1

year one

using a rule of three is possible to find how much is 8.75 od 4500 km2

Let

4500 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$4500:100\\x:8.75\\\frac{4500}{100}=\frac{x}{8.75}\\x=\frac{4500*8.75}{100} \\x=393.75\\$

at the end of the year one, the area will be

4500-393.75=4106.25

this will be the initial area for the year 2.

Step 2

repite the step 1 with area initial =4106.25 km2

4106.25 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$4106.25:100\\x:8.75\\\frac{4106.25}{100}=\frac{x}{8.75}\\x=\frac{4106.25*8.75}{100} \\x=359.29\\$

at the end of the year 2, the area will be

4106-359.29=3746.70

this will be the initial area for the year 3.

Step 3

repite the step 1 with area initial =4106.25 km2

3746.70 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$3746.70:100\\x:8.75\\\frac{3746.70}{100}=\frac{x}{8.75}\\x=\frac{3746.7*8.75}{100} \\x=327.83\\$

at the end of the year 3, the area will be

3746.70-327.83=3419.09

this will be the initial area for the year 4.

Step 4

year four

repite the step 1 with area initial =3419.09 km2

3419.09 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$3419.09:100\\x:8.75\\\frac{3419.09}{100}=\frac{x}{8.75}\\x=\frac{3419.09*8.75}{100} \\x=299.17\\$

at the end of the year 4, the area will be

3419.09-299.173=3119.82

this will be the initial area for the year 5.

Step 5

year five

repite the step 1 with area initial =3119.82 km2

3119.82 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$3119.82:100\\x:8.75\\\frac{3119.82}{100}=\frac{x}{8.75}\\x=\frac{3119.82*8.75}{100} \\x=272.99\\$

at the end of the year 5, the area will be

3119.82-272.99=2846.92

this will be the initial area for the year 6.

Step 6

year six

repite the step 1 with area initial =2846.92km2

2846.92 km2 ⇒ 100$

x?km2 ⇒8.75

do the relation and isolate x

$2846.92:100\\x:8.75\\\frac{2846.92}{100}=\frac{x}{8.75}\\x=\frac{2846.92*8.75}{100} \\x=249.10\\$

at the end of the year six, the area will be

2846.92-249.10=2597.82 square kilometers

Have a great day.

8 0

3 years ago